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Let $\mathcal{N}$ denote the set of all products of (powers of) $2,3$ and $5$: $$ \mathcal{N} = \{ 2^a 3^b 5^c \ : \ a,b,c \geq 0 \} \subset \mathbb{N}.$$ We use the elements of $\mathcal{N}$ to partition $\mathbb{N}$ into intervals. In other words, we order $\mathcal{N} = \{N_0 < N_1 < \dots \}$ and consider the intervals $I_k = [N_k,N_{k+1})$. I will call an interval $I_k$ exceptional if $N_k$ is odd. Let $E$ be the union of all exceptional intervals.

Is it true that $E$ has asymptotic density zero?

For instance: $\mathcal{N} = \{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,\dots\}$ so the few initial intervals are $I_0 = [1,2)$, $I_1 = [2,3),\ldots,I_{10} = [15,16)$, $I_{11} = [16,18)$, $I_{12} = [18,20)$, $I_{13} = [20,24)$, $I_{14} = [24,25)$, $I_{15} = [25,27)$. The intervals $[15,16)$ and $[25,27)$ are exceptional while $[16,18)$, $[18,20)$, $[20,24)$, $[24,25)$ are not exceptional. A few initial elements of $E$ are $1,3,5,9,15,25,26,27,28,29 \dots$.

Rationale: The number of points in $\mathcal{N}$ in the interval $[X,2X)$ for large $X$ is of the order of $\log^2 X$. On the other hand, the number of points in $\mathcal{N}$ that are initial points of exceptional intervals (i.e., integers of the form $3^b 5^c$) in $[X,2X)$ is of the order of $\log X$. If we knew that all of the intervals $I_k$ contained in $[X,2X)$ have roughly the same length, then clearly this would imply that $| E \cap [X,2X)| \simeq X/\log X = o(X)$, as needed. If instead of $\{2,3,5\}$ we only had $\{2,3\}$ then this type of an argument (together with the basic fact that the sequence $n \log_2 3$ is equidistributed modulo $1$) would suffice to show that $d(E) = 0$. It seems like the variant for three primes might follow from more careful analysis of expressions like $n \log_2 3 + m \log_2 5 \bmod 1$.

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  • $\begingroup$ @GHfromMO I might be going crazy, but it seems to me that the estimates in the question are correct. If I want $n = 2^a 3^b 5^c$ to be between $X$ and $2X$ then I can choose $a$ and $b$ in $O(\log X)$ ways each, and then I have $O(1)$ ways to choose $c$ (in fact, I have either $0$ or $1$ way). I think you would have been right if we were looking at $[1,X)$ instead of $[X,2X)$, maybe that's the source of the confusion? $\endgroup$
    – Jakub Konieczny
    Commented Apr 10 at 20:26
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    $\begingroup$ You are right, my bad. Tricky logarithms. I will delete my comment. $\endgroup$
    – GH from MO
    Commented Apr 10 at 20:31
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    $\begingroup$ @MaxMuller: it's possible there's other context but the "rationale" paragraph at the end seems like good enough explanation/motivation for this problem to me. $\endgroup$
    – Sam Hopkins
    Commented Apr 10 at 22:21
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    $\begingroup$ @MaxMuller: My motivation comes from a generalisation of the notion of an automatic sequence, which is tentatively dubbed "asymptotically automatic". It turns out that the asymptotic variant of Cobham's theorem is significantly more complicated than the original. If the answer to my question is positive, then one can construct an interesting (at least to me!) example of a sequence that is asymptotically automatic in bases 2, 3 and 5, which is impossible for automatic sequences. $\endgroup$
    – Jakub Konieczny
    Commented Apr 10 at 23:04
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    $\begingroup$ Since automatic sequences are a bit of an esoteric subject, and since the origin does not seem to shed much light on the problem, I did not mention them in the post. (More discussion on asymptotically automatic sequences can be found in arxiv.org/abs/2305.09885 and arxiv.org/abs/2209.09588). $\endgroup$
    – Jakub Konieczny
    Commented Apr 10 at 23:05

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